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A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)
- a)g
- b)g / √2
- c)g / √3
- d)g / √5
Correct answer is option 'C'. Can you explain this answer?
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A triangular block of mass M with angles 30°, 60° and 90° rest with i...
To solve this problem, we can use the concept of equilibrium and apply Newton's second law to both blocks.
1. Free Body Diagram of the Cubical Block:
The cubical block has a mass of m and is at rest relative to the triangular block. The forces acting on the cubical block are:
- Weight (mg) acting vertically downwards
- Normal force (N) from the triangular block acting vertically upwards
- Friction force (f) from the triangular block acting horizontally towards the left (opposite to the direction of acceleration)
2. Free Body Diagram of the Triangular Block:
The triangular block has a mass of M and is accelerating relative to the table. The forces acting on the triangular block are:
- Weight (Mg) acting vertically downwards
- Normal force (N) from the table acting vertically upwards
- Friction force (F) from the table acting horizontally towards the right (in the direction of acceleration)
3. Equations of Motion:
For the cubical block to be in equilibrium relative to the triangular block, the net force on it must be zero. Therefore, we have:
f = mg * sin(60°) [1]
For the triangular block to move with the cubical block at rest relative to it, the net force on it must be zero. Therefore, we have:
F = Mg * sin(30°) [2]
4. Relating Acceleration to Net Forces:
The acceleration (a) of the triangular block relative to the table is related to the net force (F) acting on it by:
F = Ma [3]
The acceleration (a) of the cubical block relative to the triangular block is related to the net force (f) acting on it by:
f = ma [4]
5. Solving the Equations:
Substituting equations [1] and [2] into equations [3] and [4] respectively, we obtain:
Mg * sin(30°) = Ma [5]
mg * sin(60°) = ma [6]
Dividing equation [5] by equation [6], we get:
(Mg * sin(30°))/(mg * sin(60°)) = (Ma)/(ma)
cancelling M and m:
(g * sin(30°))/(g * sin(60°)) = (a)/(a)
sin(30°)/sin(60°) = 1
(1/2)/(√3/2) = 1
1/(√3/2) = 1
2/√3 = 1
√3 = 2
√3/2 = 1
Hence, the acceleration which M must have relative to the table to keep m stationary relative to the triangular block is g/√3, which corresponds to option C.
1. Free Body Diagram of the Cubical Block:
The cubical block has a mass of m and is at rest relative to the triangular block. The forces acting on the cubical block are:
- Weight (mg) acting vertically downwards
- Normal force (N) from the triangular block acting vertically upwards
- Friction force (f) from the triangular block acting horizontally towards the left (opposite to the direction of acceleration)
2. Free Body Diagram of the Triangular Block:
The triangular block has a mass of M and is accelerating relative to the table. The forces acting on the triangular block are:
- Weight (Mg) acting vertically downwards
- Normal force (N) from the table acting vertically upwards
- Friction force (F) from the table acting horizontally towards the right (in the direction of acceleration)
3. Equations of Motion:
For the cubical block to be in equilibrium relative to the triangular block, the net force on it must be zero. Therefore, we have:
f = mg * sin(60°) [1]
For the triangular block to move with the cubical block at rest relative to it, the net force on it must be zero. Therefore, we have:
F = Mg * sin(30°) [2]
4. Relating Acceleration to Net Forces:
The acceleration (a) of the triangular block relative to the table is related to the net force (F) acting on it by:
F = Ma [3]
The acceleration (a) of the cubical block relative to the triangular block is related to the net force (f) acting on it by:
f = ma [4]
5. Solving the Equations:
Substituting equations [1] and [2] into equations [3] and [4] respectively, we obtain:
Mg * sin(30°) = Ma [5]
mg * sin(60°) = ma [6]
Dividing equation [5] by equation [6], we get:
(Mg * sin(30°))/(mg * sin(60°)) = (Ma)/(ma)
cancelling M and m:
(g * sin(30°))/(g * sin(60°)) = (a)/(a)
sin(30°)/sin(60°) = 1
(1/2)/(√3/2) = 1
1/(√3/2) = 1
2/√3 = 1
√3 = 2
√3/2 = 1
Hence, the acceleration which M must have relative to the table to keep m stationary relative to the triangular block is g/√3, which corresponds to option C.
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A triangular block of mass M with angles 30°, 60° and 90° rest with i...
Use result, a = gtan θ, a = gtan 30° = g√3
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A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer?
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A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer?.
A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A triangular block of mass M with angles 30°, 60° and 90° rest with its 30° – 90° side on a horizontal table. A cubical block of mass m rests on the 60° – 30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block is (assuming frictionless contact)a)gb)g / √2c)g / √3d)g / √5Correct answer is option 'C'. Can you explain this answer?.
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